Flat maps with geometrically regular fibres have standard smooth local presentations

Set-up. Write $P:=R[x_1,\dots ,x_n]$, $S=P\mathrm{/}I$, $P_{\kappa }:=\kappa [x_1,\dots ,x_n]$ and $I_{\kappa }=I{P}_{\kappa }$, so that $F=P_{\kappa }\mathrm{/}{I}_{\kappa }$ and the primes ${\mathfrak{q}}^{\mathrm{\prime }}\subseteq P$, $\bar{\mathfrak{q}}\subseteq F$ correspond to one another and contract to $\mathfrak{q}$; in the fibre, $\bar{\mathfrak{q}}$ corresponds to the prime ${\bar{\mathfrak{q}}}^{\mathrm{\prime }}:={\mathfrak{q}}^{\mathrm{\prime }}{P}_{\kappa }+I_{\kappa }\subseteq P_{\kappa }$ and $F\mathrm{/}\bar{\mathfrak{q}}=(S\mathrm{/}\mathfrak{q}){\otimes }_R\kappa$ has fraction field $\kappa (\mathfrak{q})$ because $S\mathrm{/}\mathfrak{q}$ is a domain.

The fibre is regular at its own residue field, so the fibre ideal has $c$ generators. Taking for $K$ the identity extension $\kappa \mathrm{/}\kappa$ in [F1], the local ring $A^{\mathrm{\prime }}:=P_{\kappa ,{\bar{\mathfrak{q}}}^{\mathrm{\prime }}}$ is regular local by [F14], since the field $\kappa$ is regular Noetherian, and has dimension $\mathrm{ht}({\bar{\mathfrak{q}}}^{\mathrm{\prime }})$ by [F12], and its quotient $F_{\bar{\mathfrak{q}}}=A^{\mathrm{\prime }}\mathrm{/}{I}_{\kappa }{A}^{\mathrm{\prime }}$ is a regular local ring by the hypothesis. Since $I_{\kappa }=(f_1,\dots ,f_m)$ and every $f_j$ lies in ${\bar{\mathfrak{q}}}^{\mathrm{\prime }}$, [F5] applies and shows that $I_{\kappa }{A}^{\mathrm{\prime }}$ is generated by an initial part of a regular system of parameters of $A^{\mathrm{\prime }}$; put $c:=\mathrm{ht}({\bar{\mathfrak{q}}}^{\mathrm{\prime }})-\dim F_{\bar{\mathfrak{q}}},$ so that $I_{\kappa }{A}^{\mathrm{\prime }}$ is generated by $c$ elements whose classes in $I_{\kappa }{A}^{\mathrm{\prime }}\mathrm{/}{\bar{\mathfrak{q}}}^{\mathrm{\prime }}(I_{\kappa }{A}^{\mathrm{\prime }})$ are $\kappa (\mathfrak{q})$-independent. The classes of the images of $f_1,\dots ,f_m$ span that $\kappa (\mathfrak{q})$-vector space of dimension $c$, so after renumbering, the images of $f_1,\dots ,f_c$ form a basis and generate $I_{\kappa }{A}^{\mathrm{\prime }}$ by [F6] applied to the finitely generated $A^{\mathrm{\prime }}$-module $I_{\kappa }{A}^{\mathrm{\prime }}$.

The $K$-rational point of the fibre and the rank computation. By [F7] choose an algebraic closure $K\mathrm{/}\kappa (\mathfrak{q})\supseteq \kappa$; it is perfect by [F7]. The composite $F{\otimes }_{\kappa }K\to \kappa (\mathfrak{q}){\otimes }_{\kappa }K\to K$, $a\otimes b\mapsto ab$, obtained from the algebraic closure $\kappa (\mathfrak{q})\subseteq K$, is a surjective $K$-algebra homomorphism (it is $K$-linear and hits $K$), and restricting it to $F$ recovers the quotient map $F\to F\mathrm{/}\bar{\mathfrak{q}}\hookrightarrow \kappa (\mathfrak{q})$; hence its kernel $Q$ is a maximal ideal of $F_K:=F{\otimes }_{\kappa }K$ with $F_K\mathrm{/}Q\cong K$, so $Q$ lies over $\bar{\mathfrak{q}}$ and its preimage $\bar{\mathfrak{n}}\subseteq K[x_1,\dots ,x_n]$ is maximal with $K[x]\mathrm{/}\bar{\mathfrak{n}}\cong K$. By [F1] applied to $K\mathrm{/}\kappa$ the local ring $(F_K{)}_Q$ is regular local; put $d:=\dim (F_K{)}_Q$.

The chosen equations also generate the ideal of the $K$-fibre. Put $K[x]:=K[x_1,\dots ,x_n]$ and $I_K:=IK[x]=I_{\kappa }K[x]$, so $F_K=K[x]\mathrm{/}{I}_K$. Since the images of $f_1,\dots ,f_c$ generate $I_{\kappa }{A}^{\mathrm{\prime }}$ by step 2.1, the cokernel $I_{\kappa }{A}^{\mathrm{\prime }}\mathrm{/}(f_1,\dots ,f_c)A^{\mathrm{\prime }}$ is zero, so $I_{K}K[x{]}_{\bar{\mathfrak{n}}}=(f_1,\dots ,f_c)K[x{]}_{\bar{\mathfrak{n}}}$ by [F10] (right exactness of base change of scalars, and localisation commuting with it). Hence $T_K:=K[x]\mathrm{/}(f_1,\dots ,f_c)$ satisfies $(T_K{)}_{\bar{\mathfrak{n}}}=K[x{]}_{\bar{\mathfrak{n}}}\mathrm{/}{I}_{K}K[x{]}_{\bar{\mathfrak{n}}}=(F_K{)}_Q$, which is regular local of dimension $d$. Now $\bar{\mathfrak{n}}$ is a maximal ideal of the finite-type $K$-domain $K[x]$, so $\mathrm{ht}(\bar{\mathfrak{n}})+\dim (K[x]\mathrm{/}\bar{\mathfrak{n}})=\dim K[x]=n$ by [F8], that is, $\mathrm{ht}(\bar{\mathfrak{n}})=n$.

The remaining relations die after inverting. Put $T:=R[x_1,\dots ,x_n]\mathrm{/}(f_1,\dots ,f_c)$ and $J:=\ker (T\to S)=I\mathrm{/}(f_1,\dots ,f_c)$; the ideal $J$ is finitely generated because $I$ is. Since $f_1,\dots ,f_c$ generate $I_{\kappa }{P}_{\kappa ,{\bar{\mathfrak{q}}}^{\mathrm{\prime }}}$ by step 2.1, the map $T_{{\mathfrak{q}}^{\mathrm{\prime }}}{\otimes }_{R_{\mathfrak{p}}}\kappa ⟶S_{\mathfrak{q}}{\otimes }_{R_{\mathfrak{p}}}\kappa ,\kappa [x{]}_{{\bar{\mathfrak{q}}}^{\mathrm{\prime }}}\mathrm{/}({\bar{f}}_1,\dots ,{\bar{f}}_c)⟶\kappa [x{]}_{{\bar{\mathfrak{q}}}^{\mathrm{\prime }}}\mathrm{/}{I}_{\kappa },$ is an isomorphism. The ring $S_{\mathfrak{q}}$ is flat over $R_{\mathfrak{p}}$ by hypothesis, so [F9] gives ${\mathrm{Tor}}_1^{R_{\mathfrak{p}}}(\kappa ,S_{\mathfrak{q}})=0$; the Tor long exact sequence of [F9] applied to $0\to J_{{\mathfrak{q}}^{\mathrm{\prime }}}\to T_{{\mathfrak{q}}^{\mathrm{\prime }}}\to S_{\mathfrak{q}}\to 0$ therefore makes $J_{{\mathfrak{q}}^{\mathrm{\prime }}}{\otimes }_{R_{\mathfrak{p}}}\kappa \to T_{{\mathfrak{q}}^{\mathrm{\prime }}}{\otimes }_{R_{\mathfrak{p}}}\kappa$ injective, while its composite with the isomorphism above is zero; hence $J_{{\mathfrak{q}}^{\mathrm{\prime }}}{\otimes }_{R_{\mathfrak{p}}}\kappa =0$, that is, $J_{{\mathfrak{q}}^{\mathrm{\prime }}}=\mathfrak{p}{J}_{{\mathfrak{q}}^{\mathrm{\prime }}}$. The Axiom of Dependent Choice assumed through [F9] is a consequence of the Axiom of Choice assumed in the statement, and is used only here.

The rank and the minor. Apply [F4] with $k:=K$ (perfect), $P:=K[x]$, $I:=(f_1,\dots ,f_c)$ — a generating set of that ideal, with the $f_j$ now viewed in $K[x]$ — $A:=T_K$ and $\mathfrak{m}:=\bar{\mathfrak{n}}\mathrm{/}(f_1,\dots ,f_c)\subseteq A$: this maximal ideal has residue field $A\mathrm{/}\mathfrak{m}=K$, a finite separable extension of $K$, and $A_{\mathfrak{m}}=(T_K{)}_{\bar{\mathfrak{n}}}=(F_K{)}_Q$ is regular local of dimension $d$ by step 3.1. The criterion gives ${\mathrm{rank}}_{K}J(\mathfrak{m})=n-\dim A_{\mathfrak{m}}=n-d=:{\mu }_K,$ where $J$ is the $c\times n$ Jacobian matrix of $f_1,\dots ,f_c$. Moreover ${\mu }_K={\dim }_K(I_{K}K[x{]}_{\bar{\mathfrak{n}}}/\bar{\mathfrak{n}}{I}_{K}K[x{]}_{\bar{\mathfrak{n}}})={\dim }_K\left(\right(I_{\kappa }{A}^{\mathrm{\prime }}\mathrm{/}{\bar{\mathfrak{q}}}^{\mathrm{\prime }}{I}_{\kappa }{A}^{\mathrm{\prime }}\left){\otimes }_{\kappa (\mathfrak{q})}K\right)\ge {\dim }_{\kappa (\mathfrak{q})}\left(I_{\kappa }{A}^{\mathrm{\prime }}\mathrm{/}{\bar{\mathfrak{q}}}^{\mathrm{\prime }}{I}_{\kappa }{A}^{\mathrm{\prime }}\right)=c,$ where the middle identification is base change of scalars by [F10] applied to the quotient $I_{\kappa }{A}^{\mathrm{\prime }}\mathrm{/}(f_1,\dots ,f_c)A^{\mathrm{\prime }}=0$ of step 2.1 and the last inequality is that the dimension of a vector space cannot drop under a field extension. Therefore ${\mathrm{rank}}_{K}J(\mathfrak{m})=c$, so some $c\times c$ minor $h$ of the Jacobian matrix $(\mathrm{\partial }{f}_j\mathrm{/}\mathrm{\partial }{x}_i)$ has nonzero image in $K[x]\mathrm{/}\bar{\mathfrak{n}}=K$.

The minor descends to $\mathfrak{q}$. By the monomial formula of [F3], partial differentiation is linear over the coefficient ring, so the image in $K[x]$ of the minor $h\in R[x_1,\dots ,x_n]$ is the corresponding minor of the images of $f_1,\dots ,f_c$; since its image in $K[x]\mathrm{/}\bar{\mathfrak{n}}$ is nonzero we get $h\mathrm{\notin }\bar{\mathfrak{n}}$, hence $h\mathrm{\notin }\bar{\mathfrak{n}}\cap R[x]={\mathfrak{q}}^{\mathrm{\prime }}$ because $\bar{\mathfrak{n}}$ lies over ${\mathfrak{q}}^{\mathrm{\prime }}$, and hence $h\mathrm{\notin }\mathfrak{q}$.

Conclusion. The ring $T_{{\mathfrak{q}}^{\mathrm{\prime }}}$ is local with maximal ideal ${\mathfrak{q}}^{\mathrm{\prime }}{T}_{{\mathfrak{q}}^{\mathrm{\prime }}}$, which contains $\mathfrak{p}{T}_{{\mathfrak{q}}^{\mathrm{\prime }}}$ because $\mathfrak{p}\subseteq {\mathfrak{q}}^{\mathrm{\prime }}$, and $J_{{\mathfrak{q}}^{\mathrm{\prime }}}$ is a finitely generated $T_{{\mathfrak{q}}^{\mathrm{\prime }}}$-module; so $J_{{\mathfrak{q}}^{\mathrm{\prime }}}=\mathfrak{p}{J}_{{\mathfrak{q}}^{\mathrm{\prime }}}$ forces $J_{{\mathfrak{q}}^{\mathrm{\prime }}}=0$ by [F6]. By [F11] there is $g\in P\setminus {\mathfrak{q}}^{\mathrm{\prime }}$, with $J_g=0$, that is $T_g\cong S_{\bar{g}}$ as $R$-algebras, where $\bar{g}$ is the image of $g$ in $S$. Since $h\mathrm{\notin }{\mathfrak{q}}^{\mathrm{\prime }}$ by step 5.1, put $u:=gh\in P\setminus {\mathfrak{q}}^{\mathrm{\prime }}$ and write $\bar{u}$ for its image in $S$. The image of $h$ is a unit in $S_{\bar{u}}$ because $u=gh$ is inverted, and $J_u=0$ because $J_g=0$; hence $S_{\bar{u}}\cong T_u=(R[x_1,\dots ,x_n]\mathrm{/}(f_1,\dots ,f_c){)}_u$ with the $c\times c$ Jacobian minor $h$ a unit, a standard smooth presentation of relative dimension $n-c$ by [F2]. Since $u\mathrm{\notin }{\mathfrak{q}}^{\mathrm{\prime }}$, its image $\bar{u}\mathrm{\notin }\mathfrak{q}$, so this chart witnesses that $R\to S$ is standard smooth at $\mathfrak{q}$.